1. Lect. 6 Pole Placement Basil Hamed
Control Systems
Lect. 6 Pole Placement
Basil Hamed

2. State-Space Model
We showed that state-space methods, like transform methods, are simply tools for analyzing and designing feedback control systems. However, state-space techniques can be applied to a wider class of systems than transform methods.
Systems with nonlinearities and multiple-input, multiple-output systems are just two of the candidates for the state-space approach
Basil Hamed

3. State-Space Model
In Chapter 9, we applied frequency domain methods to system design(PID, Lead-Lag) .
The basic design technique is to create a compensator in cascade with the plant or in the feedback path that has the correct additional poles and zeros to yield a desired transient response and steady-state error.
One of the drawbacks of frequency domain methods of design, using root locus techniques, is that after designing the location of the dominant second-order pair of poles, we keep our fingers crossed, hoping that the higher-order poles do not affect the second-order approximation.
Basil Hamed

4. State-Space Model
we would like to be able to do is specify all closed-loop poles of the higher-order system
Frequency domain methods of design do not allow us to specify all poles in systems of order higher than 2 because they do not allow for a sufficient number of unknown parameters to place all of the closed-loop poles uniquely
state-space methods do not allow the specification of closed-loop zero locations, which frequency domain methods do allow through placement of the lead compensator zero.
This is a disadvantage of state-space methods, since the location of the zero does affect the transient response
Basil Hamed

5. Controller Design
Introduce additional parameters into a system so that we can control the location of all closed-loop poles. An nth-order feedback control system has an n-th-order closed-loop characteristic equation of the form
Since the coefficient of the highest power of s is unity, there are n coefficients whose values determine the system's closed-loop pole locations.
Thus, if we can introduce n adjustable parameters into the system and relate them to the coefficients in above Eq., all of the poles of the closed-loop system can be set to any desired location.
rder closed-loop characteristic equation of the form
Basil Hamed

6. Topology for Pole Placement
a plant represented in state space by
Figure 1 State-space representation of a plant
Basil Hamed

7. Topology for Pole Placement
In a typical feedback control system, the output, y, is fed back to the summing junction.
It is now that the topology of the design changes. Instead of feeding back y, what if we feed back all of the state variables?
If each state variable is fed back to the control, u, through a gain, ki there would be n gains, ki that could be adjusted to yield the required closed-loop pole values.
Basil Hamed

8. Topology for Pole Placement
The state equations for the closed-loop system of shown Figure can be written by inspection as
Figure 2 Plant with state-variable feedback
Basil Hamed

9. Topology for Pole Placement
Before continuing, we should have a good idea of how the feedback system of Figure 2 is actually implemented.
As an example, assume a plant signal-flow graph in phase-variable form, as shown in Figure 3(a). Each state variable is then fed back to the plant's input, u, through a gain, ki, as shown in Figure 3(b).
Basil Hamed

10. Topology for Pole Placement
FIGURE 3 a. Phase variable representation for plant; b. plant with state variable feedback
Basil Hamed

11. Controllability
Consider the parallel form shown in Figure 4(a).To control the pole location of the closed-loop system, we are saying implicitly that the control signal, u, can control the behavior of each state variable in x.
If any one of the state variables cannot be controlled by the control u, then we cannot place the poles of the system where we desire.
For example, in Figure 4(b), if x1 were not controllable by the control signal and if x1 also exhibited an unstable response due to a nonzero initial condition,
There would be no way to effect a state-feedback design to stabilize x1; x1 would perform in its own way regardless of the control signal, u. Thus, in some systems, a state-feedback design is not possible.
Basil Hamed

12. Controllability
FIGURE 4 Comparison of a. controllable and b. uncontrollable systems
Basil Hamed

13. Controllability
If an input to a system can be found that takes every state variable from a desired initial state to a desired final state, the system is said to be controllable; otherwise, the system is uncontrollable. Pole placement is a viable design technique only for systems that are controllable.
Basil Hamed

14. Controllability by Inspection
We can explore controllability from another viewpoint: that of the state equation itself. When the system matrix is diagonal, as it is for the parallel form, it is apparent whether or not the system is controllable. For example, the state equation for Figure 4(a) is
Basil Hamed

15. Controllability by Inspection
Since each of above Eqs. is independent and decoupled from the rest, the control u affects each of the state variables. This is controllability from another perspective. Now let us look at the state equations for the system of Figure(b):
From the state equations above, we see that state variable X1 is not controlled by the control u. Thus, the system is said to be uncontrollable
In summary, a system with distinct eigenvalues and a diagonal system matrix is controllable if the input coupling matrix B does not have any rows that are zero.
Basil Hamed

16. The Controllability Matrix
An nth-order plant whose state equation is
is of rank n, where CM is called the controllability matrix
Basil Hamed

17. The Controllability Matrix
PROBLEM: Given the system of Figure below, represented by a signal-flow diagram, determine its controllability.
SOLUTION: The state equation for the system written from the signal-flow diagram is
Basil Hamed

18. The Controllability Matrix
The controllability matrix is
The rank of CM equals the number of linearly independent rows or columns. The rank can be found by finding the highest-order square submatrix that is nonsingular. The determinant of CM = - 1
Since the determinant is not zero, the 3x3 matrix is nonsingular, and the rank of CM is 3
We conclude that the system is controllable since the rank of CM equals the system order.
Basil Hamed

19. Pole Placement for Plants in Phase-Variable Form
To apply pole-placement methodology to plants represented in phase-variable form, we take the following steps:
Represent the plant in phase-variable form.
Feed back each phase variable to the input of the plant through a gain, ki.
Find the characteristic equation for the closed-loop system represented in Step 2.
Decide upon all closed-loop pole locations and determine an equivalent characteristic equation.
Equate like coefficients of the characteristic equations from Steps 3 and 4 and solve for ki.
Basil Hamed

20. Pole Placement for Plants in Phase-Variable Form
Following these steps, the phase-variable representation of the plant is given by
The characteristic equation of the plant is thus
Now form the closed-loop system by feeding back each state variable to u, forming
U=-Kx Where
The ki’s are the phase variables' feedback gains
Basil Hamed

21. Pole Placement for Plants in Phase-Variable Form
Since above matrix is in phase-variable form, the characteristic equation of the closedloop system can be written by inspection as
Basil Hamed

22. Pole Placement for Plants in Phase-Variable Form
Now assume that the desired characteristic equation for proper pole placement is
where the di’s are the desired coefficients
from which
Now that we have found the denominator of the closed-loop transfer function, let us find the numerator. For systems represented in phase-variable form, we learned that the numerator polynomial is formed from the coefficients of the output coupling matrix, C. Since Figures 3(a) and (b) are both in phase-variable form and have the same output coupling matrix, we conclude that the numerators of their transfer functions are the same
Basil Hamed

23. Controller Design for Phase-Variable Form
PROBLEM: Given the plant
design the phase-variable feedback gains to yield 9.5% overshoot and a settling time of 0.74 second.
SOLUTION:
We begin by calculating the desired closed-loop characteristic equation. Using the transient response requirements, the closed-loop poles are ± j7.2.
Since the system is third-order, we must select another closed-loop pole. The closed-loop system will have a zero at —5, the same as the open-loop system. We could select the third closed-loop pole to cancel the closed-loop zero
let us choose -5.1 as the location of the third closed-loop pole.
Basil Hamed

24. Controller Design for Phase-Variable Form
Now draw the signal-flow diagram for the plant. The result is shown in Figure below(a). Next feed back all state variables to the control, u, through gains ki, as shown in Figure below(b).
Basil Hamed

25. Controller Design for Phase-Variable Form
Writing the closed-loop system's state equations
we identify the closed-loop system matrix as
Basil Hamed

26. Controller Design for Phase-Variable Form
To find the closed-loop system's characteristic equation, form This equation must match the desired characteristic equation formed from the poles j7.2, j7.2, and -5.1, which were previously determined. Equating the coefficients of above Eqs., we obtain
Basil Hamed

27. Controller Design for Phase-Variable Form
we obtain the following state-space representation of the closed-loop system:
The transfer function is
Basil Hamed

28. Controller Design for Phase-Variable Form
Simulation of closed-loop system
a simulation of the closed-loop system, shows 11.5% overshoot and a settling time of 0.8 second. A redesign with the third pole canceling the zero at —5 will yield performance equal to the requirements
Basil Hamed

29. Alternative Approaches to Controller Design
We showed how to design state-variable feedback to yield desired closed loop poles.
We demonstrated this method using systems represented in phase variable form and saw how simple it was to calculate the feedback gains.
Many times the physics of the problem requires feedback from state variables that are not phase variables.
For these systems we have some choices for a design methodology
Basil Hamed

30. Alternative Approaches to Controller Design
Controllability Canonical Form (CCF) The characteristic equation of A is The dynamic equations in Eqs. Shown above are transformed into CCF: Where
Basil Hamed

31. Alternative Approaches to Controller Design
Then,
Basil Hamed

32. Alternative Approaches to Controller Design
Example Consider the coefficient matrices of the state equations The state equations are to be transformed to CCF. The characteristic equation of A is
Basil Hamed

33. Alternative Approaches to Controller Design
We can show that S is nonsingular, so the system can be transformed into the CCF. Substituting S and M into
Thus, the CCF model is given by
Basil Hamed

34. Observability
Recall that the ability to control all of the state variables is a requirement for the design of a controller. State-variable feedback gains cannot be designed if any state variable is uncontrollable. Uncontrollability can be viewed best with diagonalized systems. The signal-flow graph showed clearly that the uncontrollable state variable was not connected to the control signal of the system.
A similar concept governs our ability to create a design for an observer. Specifically, we are using the output of a system to deduce the state variables. If any state variable has no effect upon the output, then we cannot evaluate this state variable by observing the output
Basil Hamed

35. Observability
The ability to observe a state variable from the output is best seen from the diagonalized system. Figure(a) shows a system where each state variable can be observed at the output since each is connected to the output. Figure(b) is an example of a system where all state variables cannot be observed at the output. Here X1 is not connected to the output and could not be estimated from a measurement of the output. We now make the following definition based upon the previous discussion:
Basil Hamed

36. Observability
If the initial-state vector, x(t0), can be found from u(t) and y(t) measured over a finite interval of time from t0, the system is said to be observable; otherwise the system is said to be unobservable
Simply stated, observability is the ability to deduce the state variables from a knowledge of the input, u(t), and the output, y(t).
Pole placement for an observer is a viable design technique only for systems that are observable
Basil Hamed

37. Observability
Observability by Inspection We can also explore observability from the output equation of a diagonalized system. The output equation for the diagonalized system of Figure (a) is y = Cx=[1 1 1]x (12.76) On the other hand, the output equation for the unobservable system of Figure (b) is y = Cx=[0 1 1]x (12.77) Notice that the first column of Eq. (12.77) is zero. For systems represented in parallel form with distinct eigenvalues, if any column of the output coupling matrix is zero, the diagonal system is not observable.
Basil Hamed

38. Observability
The Observability Matrix An nth-order plant whose state and output equations are, respectively, x = Ax + Bu y = Cx is completely observable(rank is n) if the matrix
Basil Hamed

39. Observability
Example Observability via the Observability Matrix PROBLEM: Determine if the system of Figure below is observable.
Basil Hamed

40. Observability
SOLUTION: The state and output equations for the system are
Thus, the observability matrix, OM, is
Since the determinant of OM equals —344, OM is of full rank equal to 3. The
system is thus observable.
Basil Hamed

41. Observability
Remark
You might have been misled and concluded by inspection that the system is unobservable because the state variable x1 is not fed directly to the output. Remember that conclusions about observability by inspection are valid only for diagonalized systems that have distinct eigenvalues
Basil Hamed

42. Observability
PROBLEM: Determine whether the system of Figure shown is observable.
Basil Hamed

43. Observability
SOLUTION: The state and output equations for the system are
The observability matrix, OM, for this system is
The determinant for this observability matrix equals 0. Thus, the observability matrix does not have full rank, and the system is not observable.
Basil Hamed

44. Observer Design
Controller design relies upon access to the state variables for feedback through adjustable gains.
This access can be provided by hardware. For example, gyros can measure position and velocity on a space vehicle.
Sometimes it is impractical to use this hardware for reasons of cost, accuracy, or availability.
In other applications, some of the state variables may not be available at all, or it is too costly to measure them or send them to the controller.
If the state variables are not available because of system configuration or cost, it is possible to estimate the states.
Estimated states, rather than actual states, are then fed to the controller
Basil Hamed

45. Observer Design
One scheme of observer design is shown in below. An observer, sometimes called an estimator, is used to calculate state variables that are not accessible from the plant. Here the observer is a model of the plant.
Basil Hamed

46. Observer Design
Let us look at the disadvantages of such a configuration. Assume a plant,
Basil Hamed

47. Observer Design
Thus, the dynamics of the difference between the actual and estimated states is unforced, and if the plant is stable, this difference, due to differences in initial state vectors, approaches zero.
However, the speed of convergence between the actual state and the estimated state is the same as the transient response of the plant since the characteristic equation for Eq. (12.59a) is the same as that for Eq. (12.57a).
Since the convergence is too slow, we seek a way to speed up the observer and make its response time much faster than that of the controlled closed-loop system, so that, effectively, the controller will receive the estimated states instantaneously.
Basil Hamed

48. Observer Design
To increase the speed of convergence between the actual and estimated states, we use feedback, shown conceptually in Figure below
Basil Hamed

49. Observer Design
and in more detail in Figure
Basil Hamed

50. Observer Design
The error between the outputs of the plant and the observer is fed back to the derivatives of the observer's states. The system corrects to drive this error to zero. With feedback we can design a desired transient response into the observer that is much quicker than that of the plant or controlled closed-loop system.
When we implemented the controller, we found that the phase-variable or controller canonical form yielded an easy solution for the controller gains. In designing an observer, it is the observer canonical form that yields the easy solution for the observer gains.
Basil Hamed

51. Observer Design
Figure (a) shows an example of a third-order plant represented in observer canonical form. In Figure(b), the plant is configured as an observer with the addition of feedback, as previously described
Basil Hamed

52. Observer Design
The design of the observer is separate from the design of the controller. Similar to the design of the controller vector, K, the design of the observer consists of evaluating the constant vector, L, so that the transient response of the observer is faster than the response of the controlled loop in order to yield a rapidly updated estimate of the state vector. We now derive the design methodology.
We will first find the state equations for the error between the actual state vector and the estimated state vector, (x - ẋ). Then we will find the characteristic equation for the error system and evaluate the required L to meet a rapid transient response for the observer.
Then we will find the characteristic equation for the error system and evaluate the required L to meet a rapid transient response for the observer.
Basil Hamed

53. Observer Design
Writing the state equations of the observer from Figure, we have
Basil Hamed

54. Observer Design Subtracting Eqs. (12.60) from (12.61), we obtain
where x - ẋ is the error between the actual state vector and the estimated state vector, and y-ŷ is the error between the actual output and the estimated out-put. Substituting the output equation into the state equation, we obtain the state equation for the error between the estimated state vector and the actual state vector:
Basil Hamed

55. Observer Design
Equation (12.64a) is unforced. If the eigenvalues are all negative, the estimated state vector error, ex, will decay to zero. The design then consists of solving for the values of L to yield a desired characteristic equation or response for Eqs. (12.64). The characteristic equation is found from Eqs. (12.64) to be Now we select the eigenvalues of the observer to yield stability and a desired transient response that is faster than the controlled closed-loop response. These eigenvalues determine a characteristic equation that we set equal to Eq. (12.65) to solve for L.
Basil Hamed

56. Observer Design
Let us demonstrate the procedure for an nth-order plant represented in observer canonical form. We first evaluate A — LC. The form of A, L, and C can be derived by extrapolating the form of these matrices from a third-order plant. Thus,
Basil Hamed

57. Observer Design The characteristic equation for A - LC is
Notice the relationship between Eq. (12.67) and the characteristic equation, det(sI - A) = 0, for the plant, which is
Thus, if desired, Eq. (12.67) can be written by inspection if the plant is represented in observer canonical form. We now equate Eq. (12.67) with the desired closed-loop observer characteristic equation, which is chosen on the basis of a desired transient response. Assume the desired characteristic equation is
Basil Hamed

58. Observer Design
We can now solve for the li's by equating the coefficients of Eqs. (12.67) and (12.69):
Example Observer Design for Observer Canonical Form
PROBLEM: Design an observer for the plant
which is represented in observer canonical form. The observer will respond 10 times faster than the controlled loop designed in previous Example
For a 20.8% overshoot and a settling time of 4 seconds, a factor of the characteristic equation of the designed closed-loop system is 𝑠 2 +2𝑠+5 .
Basil Hamed

59. Observer Design
SOLUTION: 1. First represent the estimated plant in observer canonical form. The result is shown in Figure (a).
Basil Hamed

60. Observer Design
2. Now form the difference between the plant's actual output, y, and the observer's estimated output, y, and add the feedback paths from this difference to the derivative of each state variable. The result is shown in Figure (b).
Basil Hamed

61. Observer Design
Next find the characteristic polynomial. The state equations for the estimated plant shown in Figure(a) are
From Eqs. (12.64) and (12.66), the observer error is
Basil Hamed

62. Observer Design
Using Eq. (12.65), we obtain the characteristic polynomial
4. Now evaluate the desired polynomial, set the coefficients equal to those of Eq. (12.74), and solve for the gains, li. From Eq. (12.50), the closed-loop controlled system has dominant second-order poles at - 1 ±j2. To make our observer 10 times faster, we design the observer poles to be at -10 ±j20. We select the third pole to be 10 times the real part of the dominant second-order
poles, or —100. Hence, the desired characteristic polynomial is
Basil Hamed

63. Observer Design
Simulation showing response of observer: a. closed-loop; b. open-loop with observer gains disconnected
Basil Hamed

64. Observer Design
Since the dominant pole of the observer is -10 ±/20, the expected settling time should be about 0.4 second. It is interesting to note the slower response in Figure (b), where the observer gains are disconnected, and the observer is simply a copy of the plant with a different initial condition.
Basil Hamed

65. Steady-State Error Design via Integral Control
In Section 7.8, we discussed how to analyze systems represented in state space for steady-state error. In this section, we discuss how to design systems represented in state space for steady-state error.
Consider Figure shown. The previously designed controller discussed is shown inside the dashed box. A feedback path from the output has been added to form the error, e, which is fed forward to the controlled plant via an integrator. The integrator increases the system type and reduces the previous finite error to zero.
Basil Hamed

66. Steady-State Error Design via Integral Control
now derive the form of the state equations for the system of shown Figure and then use that form to design a controller. Thus, we will be able to design a system for zero steady-state error for a step input as well as design the desired transient response. An additional state variable, XN, has been added at the output of the leftmost integrator. The error is the derivative of this variable.
Writing the state equations from previous slide Figure, we have
Basil Hamed

67. Steady-State Error Design via Integral Control
Eqs. (12.112) can be written as augmented vectors and matrices. Hence,
But
Basil Hamed

68. Steady-State Error Design via Integral Control
Substituting Eq. (12.114) into (12.113a) and simplifying, we obtain
Thus, the system type has been increased, and we can use the characteristic equation associated with Eq. (12.115a) to design K and Ke to yield the desired transient response.
Realize, we now have an additional pole to place. The effect on the transient response of any closed-loop zeros in the final design must also be taken into consideration.
Basil Hamed

69. Steady-State Error Design via Integral Control
Example Design of Integral Control PROBLEM: Consider the plant of Eqs. (12.116):
Design a controller without integral control to yield a 10% overshoot and a settling time of 0.5 second. Evaluate the steady-state error for a unit step input.
Repeat the design of (a) using integral control. Evaluate the steady-state error for a unit step input
Basil Hamed

70. Steady-State Error Design via Integral Control
SOLUTION:
Using the requirements for settling time and percent overshoot, we find that the desired characteristic polynomial is
Since the plant is represented in phase-variable form, the characteristic polynomial for the controlled plant with state-variable feedback is
Equating the coefficients of Eqs. (12.117) and (12.118), we have
Basil Hamed

71. Steady-State Error Design via Integral Control
From previous Eqs., the controlled plant with state-variable feedback represented in phase-variable form is
Using Eq. (7.96), we find that the steady-state error for a step input is
Basil Hamed

72. Steady-State Error Design via Integral Control
b. We now use Eqs. (12.115) to represent the integral-controlled plant as follows:
Basil Hamed

73. Steady-State Error Design via Integral Control
Using Eq. (3.73) and the plant of Eqs. (12.116), we find that the transfer function of the plant is G(s) = l/(s2 + 5s + 3). The desired characteristic polynomial for the closed-loop integral-controlled system is shown in Eq. (12.117).
Since the plant has no zeros, we assume no zeros for the closed-loop system and augment Eq. (12.117) with a third pole, (s + 100), which has a real part greater than five times that of the desired dominant second-order poles. The desired third-order closed-loop system characteristic polynomial is
Basil Hamed

74. Steady-State Error Design via Integral Control
The characteristic polynomial for the system of Eqs. (12.112) is
Matching coefficients from Eqs. (12.123) and (12.124), we obtain
Basil Hamed

75. Steady-State Error Design via Integral Control
Substituting these values into Eqs. (12.122) yields this closed-loop integral controlled system:
In order to check our assumption for the zero, we now apply Eq. (3.73) to Eqs. (12.126) and find the closed-loop transfer function to be
Basil Hamed

76. Steady-State Error Design via Integral Control
Since the transfer function matches our design, we have the desired transient response.
Now let us find the steady-state error for a unit step input. Applying Eq. (7.96) to Eqs. (12.126), we obtain
Thus, the system behaves like a Type 1 system.
Basil Hamed

77. HW # , 12.14, 12.20, 12.32
Basil Hamed